EXPLANATION OF HOLMIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 18 , 2015 ' ' ' INTRODUCTION' Holmium is a chemical element with symbol Ho and atomic number 67.However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Holmium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25p6 4f 116s2 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of holmium (from (E1 to E4 ) are the following: E1 = 6 , E2 = 11.8, E3 = 22.84, and E4 = 42.5 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my new paper of 2008. ' ' EXPLANATION OF E1 = 6 eV = -E(6s2) + E(6s1) Here the E(6s2) represents the binding energy of 6s2, while the E(6s1) represents the binding energy of 6s1. The charges (-65e) of the 65 electrons of the following electron configuration (1s22s22p63s23p63d104s2 4p64d105s25p64f11) screen the nuclear charge (+ 67e) and for a perfect screening we would have ζ = 2. However the electrons of 6s penetrate the electron cloud of 4f and lead to the deformation of electron clouds. Thus ζ > 2 . Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6s1 consists of one electron, we apply the Bohr formula as E(6s1) = (-13.6057)ζ2/n2 Therefore E1 = 6 eV = -E(6s2) + E(6s1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 211.9 = 0 Then solving for ζ we get ζ = 4.62 > 2 . Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of avibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . ' ' EXPLANATION OF E2 = 11.8 eV = -E(6s1) As in the case of E1 the charges (-65e) of 65 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f11) screen the nuclear charge (+67e) and for a perfect screening we would have the same effective ζ = 2. However the one electron of 6s1 penetrates the 4f11 and leads to the deformation of electron clouds. Thus we have the same ζ > 2. Here the E(6s1) represents the binding energy of (6s1) given by applying the Bohr formula as E2 = 11.8 eV = -Ε(6s1) = - ( -13.6057)ζ2 /62 Then solving for ζ we get ζ = 5.59 > 2 . Here the ζ = 5.59 > 4.62 > 2 means that after the ionization the one electron of 6s breaks more the symmetry and leads to a great deformation of electron clouds. ' ' EXPLANATION OF E3 = 22.84 eV = -E(4f2) + E(4f1) According to the experiments the electrons of 4f11 belong to the same energy level of 6s2 with n = 6 . It consist of four pairs ( eight electrons of 4f2 , 4f2 , 4f2, and 4f2 with opposite spin) and of three electrons with parallel spin. Here the E(4f2) represents the binding energy of the first 4f2, while the E(4f1) represents the binding energy of 4f1. The charges (-54e) of the 54 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6) screen the nuclear charge (+67e) and for a perfect screening we would have ζ = 13. However in the absence of 6s2 the electrons of 4f penetrate strongly the 5p6 and provide an effective ζ , whose the value is less than the perfect screening with ζ = 13. Thus ζ < 13 . Note that the first 4f2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(4f2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 4f1 consists of one electron, we apply the Bohr formula as E(4f1) = (-13.6057)ζ2/n2 Therefore E3 = 22.84 eV = -E(4f2) + E(4f1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 818.14 = 0 Then solving for ζ we get ζ = 8.4 < 13 .' ' ' ' EXPLANATION OF E4 = 42.5 eV = -E(4f2) + E(4f1) Here the E(4f2) represents the binding energy of the second 4f2, while the E(4f1) represents the binding energy of 4f1. As in the case of E3 the charges (-54e) of the 54 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6) screen the nuclear charge (+67e) and for a perfect screening we would have ζ = 13. However in the absence of 6s2 and after the third ionization the new electron cloud of 4f penetrates also strongly the 5p6 . Thus it provides an effective ζ, whose the value is less than the perfect screening with ζ = 13. Note that the second 4f2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(4f2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 4f1 consists of one electron, we apply the Bohr formula as E(4f1) = (-13.6057)ζ2/n2 Therefore E4 = 42.5 eV = -E(4f2) + E(4f1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1525.9 = 0 Then solving for ζ we get ζ = 11.23 < 13 . Category:Fundamental physics concepts